SATHEE: Quantitative Aptitude Ques 788

Quantitative Aptitude Ques 788

Question: If $ A+B=90{}^\circ , $ then the value of $ \frac{2({{\sin }^{2}}A+{{\sin }^{2}}B)}{cose{c^{2}}(A+B)} $ is

Options:

A) $ \frac{1}{4} $

B) $ \frac{1}{2} $

C) $ 2 $

D) $ 1 $

Show Answer

Answer:

Correct Answer: C

Solution:

$ A+B=90{}^\circ $
$ \Rightarrow $ $ A=(90{}^\circ -B) $

$ \therefore $ $ \frac{2[{{\sin }^{2}}A+{{\sin }^{2}}(90{}^\circ -A)]}{cose{c^{2}}(90{}^\circ )}=\frac{2({{\sin }^{2}}A+{{\cos }^{2}}A)}{cose{c^{2}}90{}^\circ } $ $ =\frac{2}{1}=2 $

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Quantitative Aptitude Ques 788

Question: If $ A+B=90{}^\circ , $ then the value of $ \frac{2({{\sin }^{2}}A+{{\sin }^{2}}B)}{cose{c^{2}}(A+B)} $ is

Options:

Answer:

Solution:

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